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$Φ_{1} (0) = - 1$ and $Φ_{n} (0) = 1$ for $n \geq 2$

Statement

For the cyclotomic polynomials of The cyclotomic polynomials $Φ_{n} \in Z [t]$ , defined by $\prod_{d ∣ n} Φ_{d} = t^{n} - 1$ ,

$Φ_{1} (0) = - 1, Φ_{n} (0) = 1 for every n \geq 2.$

Facts & Assumptions

Given: The cyclotomic polynomials $Φ_{n} \in Z [t]$ and their defining identity; evaluation at $0$ is as in Evaluation and roots of a polynomial in a commutative target ring.

[L1]

For every $n \geq 1$ one has $\prod_{d ∣ n} Φ_{d} = t^{n} - 1$ , the product being over the positive divisors of $n$ (The recursion defines a unique monic $Φ_{n} \in Z [t]$ , of degree $φ (n)$ , The sum $\sum_{i \in S} a_{i}$ over a finite index set, and its product form, Divisibility in $Z$ : $d ∣ a$ when $a = d q$ for some integer $q$ ).

[L2]

Evaluation at an element of a commutative ring is a unital ring homomorphism $Z [t] \to Z$ (Universal property of $R [x]$ : a coefficient homomorphism and the image of $x$ determine a unique ring homomorphism), so it carries finite products to finite products.

Proof

technique · induction

1.1

$Φ_{1} = t - 1$ by The cyclotomic polynomials $Φ_{n} \in Z [t]$ , defined by $\prod_{d ∣ n} Φ_{d} = t^{n} - 1$ , so $Φ_{1} (0) = - 1$ ; this is also the base of the induction below, taken at $n = 2$ , where [L1] and [L2] give $Φ_{1} (0) Φ_{2} (0) = 0^{2} - 1 = - 1$ , hence $- Φ_{2} (0) = - 1$ and $Φ_{2} (0) = 1$ .

baseL1L2given

1.2

Inductive hypothesis: fix $n \geq 3$ and assume $Φ_{m} (0) = 1$ for every $m$ with $2 \leq m < n$ .

2.1

Evaluating the identity of [L1] at $0$ and using [L2] gives $\prod_{d ∣ n} Φ_{d} (0) = - 1$ ; separating the factor $d = 1$ , which is $- 1$ by step 1.1, leaves $\prod_{d ∣ n, d > 1} Φ_{d} (0) = 1$ .

step 1.1L1L2

3.1

Every factor with $1 < d < n$ equals $1$ by step 1.2, so the product reduces to $Φ_{n} (0) = 1$ , which is the assertion at $n$ ; the induction is complete and $Φ_{n} (0) = 1$ for every $n \geq 2$ .

step 1.2step 2.1discharge-induction∎

Remarks

What this is for. The value $1$ at the origin makes $Φ_{n} (x) \equiv 1$ modulo $x$ for every integer $x$ , so no prime dividing $x$ divides $Φ_{n} (x)$ . That is the whole mechanism behind For every $n \geq 1$ there are infinitely many primes $p$ with $p \equiv 1 (m o d n)$ , and the exceptional value at $n = 1$ is why that theorem treats $n = 1$ separately.

Depends on

Used by

For every n≥1 there are infinitely many primes p with p≡1 (mod n) Theorem

Dependency tree · two levels

40 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

P. L. Clark, Field Theory (course notes/monograph), Exercise 9.8 (standard reference, not scraped)
K. Conrad, Cyclotomic Extensions (expository blurb), Section 5 (standard reference, not scraped)

Φ1(0)=−1Φ1​(0)=−1 and Φn(0)=1Φn​(0)=1 for n≥2n≥2